i. A x = b Ax=b A x = b is inconsistent (i.e., no solution exists) if and only if rank [ A ] < rank [ A ∣ b ] . \text{rank}[A]<\text{rank}[A|b]. rank [ A ] < rank [ A ∣ b ] .
ii. A x = b Ax=b A x = b has an unique solution if and only if rank [ A ] = rank [ A ∣ b ] = n . \text{rank}[A]=\text{rank}[A|b]=n. rank [ A ] = rank [ A ∣ b ] = n .
iii. A x = b Ax=b A x = b has infinitely many solutions if and only if rank [ A ] = rank [ A ∣ b ] < n . \text{rank}[A]=\text{rank}[A|b]<n. rank [ A ] = rank [ A ∣ b ] < n .
A = [ 0 4 1 12 − 5 − 3 − 6 0 4 ] A=\begin{bmatrix}
0 & 4 & 1 \\
12 & -5 & -3 \\
-6 & 0 & 4
\end{bmatrix} A = ⎣ ⎡ 0 12 − 6 4 − 5 0 1 − 3 4 ⎦ ⎤ Swap rows 1and 2
[ 12 − 5 − 3 0 4 1 − 6 0 4 ] \begin{bmatrix}
12 & -5 & -3 \\
0 & 4 & 1 \\
-6 & 0 & 4
\end{bmatrix} ⎣ ⎡ 12 0 − 6 − 5 4 0 − 3 1 4 ⎦ ⎤ R 3 = R 3 + ( 1 / 2 ) R 1 R_3=R_3+(1/2)R_1 R 3 = R 3 + ( 1/2 ) R 1
[ 12 − 5 − 3 0 4 1 0 − 5 / 2 5 / 2 ] \begin{bmatrix}
12 & -5 & -3 \\
0 & 4 & 1 \\
0 & -5/2 & 5/2
\end{bmatrix} ⎣ ⎡ 12 0 0 − 5 4 − 5/2 − 3 1 5/2 ⎦ ⎤ R 3 = R 3 + ( 5 / 8 ) R 2 R_3=R_3+(5/8)R_2 R 3 = R 3 + ( 5/8 ) R 2
The rank of a matrix is the number of nonzero rows in the reduced matrix, so the rank is 3.
Since rank [ A ] = rank [ A ∣ b ] = 3 = n , \text{rank}[A]=\text{rank}[A|b]=3=n, rank [ A ] = rank [ A ∣ b ] = 3 = n , then A x = b Ax=b A x = b has an unique solution.
Therefore, the given system of equations is consistent.
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